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Homework Help: Sniper question not working out

  1. Feb 26, 2006 #1


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    hi... im sure all u guys have heard of the sniper question my fellow classmates have posted... but for some reason my method is not working out... i am off by a very miniscule amount which i cannot seem to find out why... any ways this is what i came up with

    Scenario: you are 1500m away from a building which is 40m high. you have a sniper rifle (in my case, with a muzzle velocity of 854m/s) and you want to shoot an object right at the edge of the 40m high building. with what angle must u adjust your sniper rifle to hit the target?

    so this is what i did... a bit diff from the others that have posted.

    first i broke my horizontal(x) and vertical(y) components up and got:
    using pythagorean theorem:
    x = root(729316 -y^2)
    y = root(729316 -x^2)

    then, since x constant velocity i solved:
    x = 1500/t

    i subbed x into y eq'n and got:

    y = root[729316 - (1500/t)^2]
    y = root[(729316t^2 - 2250000)/t^2]

    then using my kinematics formula d=v1t + 1/2at^2 i solved for the time it would take the bullet to hit the object using vertical component as my v1.

    40 = yt -4.9t^2

    i subbed y = root[(729316t^2 - 2250000)/t^2], into the equation:

    40 = [(729316t^2 - 2250000)/t^2]t - 4.9t^2
    1600 = 729316t^2 - 2250000 - 24.01t^4
    24.01t^4 - 729316t^2 + 2251600 = 0

    i solved for my realistic roots

    t = 1.757 and t = 174.277 (i highly doubt it is the second one though)

    now that i know what time equals i solved for the horizontal velocity using

    v = d/t
    x = 1500/1.757
    x = 853.7m/s

    then using this horizontal componant and the actual component (854m/s) i solve for the missing angle

    853.7 = 854cosr (r = missing angle)
    r = 1.52 degrees.

    then i found my vertical velocity using y = 854sinr

    y = 22.65m/s

    then, back to the kinematics eq'n, with all my variables solved for now, i wanted to see if my values would give me my 40m that was required.

    d=22.65(1.757) + 0.5(-9.8)(1.757)^2
    d=24.67 which is not the desired height.

    can someone please tell me where i went wrong here?

    i know my explanation is not too great but i will try my best to explain what i am trying to do. any help would be appreciated. thanks

    - Tu
  2. jcsd
  3. Feb 26, 2006 #2
    For starters, what are you doing with the Pythagorean Theorem? You don't need it.

    What information do you have? Define the sniper as the origin. I'm going to set +x to the right and +y upward. The target is at (1500m, 40m).

    Now, in the x direction we know there is no acceleration. So we have:
    [tex]x=x_0 + v_{0x}t[/tex]
    What's v0x? It's the initial velocity component in the x direction of the bullet. We know that the initial speed of the bullet is 854 m/s, so in the x direction this becomes: [tex]v_{0x}=854cos \theta[/tex] where theta is the angle above the x axis that the gun is propped up to.

    The x equation gives us one equation in two unknowns, so we can't solve this yet. So let's look at the y direction.

    To save time I'll tell you the equation we need:
    We can find v0y the same way we found v0x, in this case using the sine function. What is ay? Well, the only acceleration in the y direction here is the acceleration due to gravity, which is downward, so ay=-g.

    To sum up at this point, we have two equations in the unknowns t and theta:
    [tex]1500=854cos\theta \, t[/tex]
    [tex]40=854sin\theta \,t-4.9t^2[/tex]

    How do we solve this? My suggestion is to solve the x equation for cosine theta and substitute this into the y equation. We can do that via the relationship [tex]sin^2\theta+cos^2\theta=1[/tex]. Just plug the cosine expression into this equation, solve for the sine function, and plug sine into the y equation.

    This will result in a "biquadratic" equation for t: [tex]at^4+bt^2+c=0[/tex]. You can solve for t^2 by using the quadratic equation, and then take the square root. There will be 4 solutions.

    Two solutions will be entirely unphysical. The remaining two you can back-substitute into your equations to find the angles. Then you need to decide which angle to use.

    Please, feel free to share this!

  4. Feb 26, 2006 #3


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    ok... ill try what u said Dan... thnx alot for your help. ill post again if anything comes up.

    - Tu
  5. Feb 26, 2006 #4


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    hey... dan, i did what u said and it brought me back to where i started... the biquadratic ur method gave me was the exact same as the biquadratic method i came up with...
    24.01t^4 - 729316t^2 + 2251600 = 0
    ... however... i used ur way of finding the angle (subbing t int 40 = 854sin...) and got the correct angle. the way i did it earlier, using the same values didnt give me that answer though. which makes no sense... i plugged t into the constant horizontal velocity formula
    1.757 = 1500/x... i took that x and subbed it into
    x = 854costheta... but it gave me a diff. angle.
    is there something wrong with this approach? too much rounding maybe?

    - Tu
    Last edited: Feb 26, 2006
  6. Feb 27, 2006 #5
    As I don't have your numbers I can't tell. But I will tell you that there ARE two solutions to the problem and if you add the angles, they add to 90 degrees.

  7. Sep 16, 2007 #6
    Think about trig.

    Draw a right angle triangle.
    Put right angle bottom right
    Put 1500m on the base
    Put 40m on the vertical
    Put unkown in bottom left angle

    you have the measurements of the sides opposite and adjacent to the angle you want.

    So the trig rules say you use the Tan formulae.

    To work out the angle, you do the

    trig rule^-1(opposite/adjacent)

    which in this case means the answer is

    Tan^-1(40/1500) = 1.527525442
    to 5 s.f. = 1.5275

    hope that helps:biggrin:
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